# question for Dr.Bob22

The mole fraction of acetic acid (CH3COOH) in an aqueous solution is 0.675. Calculate the molarity of the acetic acid, if the density of the solution is 1.0266 g mL-1.

To give a different take on this, here is another way of approaching the problem. I don't claim it is better or easier. Let X = mols acetic acid; Y = mols H2O. (Two equations and two unknowns).
mol fraction acetic acid = X/(X+Y) = 0.675
solve for X in terms of Y. I have
X = 2.077*Y

(second equation)
mol fraction water = 1-X = 0.325 = Y/(X+Y)
solve for Y in terms of X. I have
0.481*X
Plug 0.481X in for Y and I get 1 mol for X; plug this value into the second equation to obtain Y = 0.481 mol.

Then 1 mol X = 1 mol CH3COOH = 60 g (You should check this as I estimated.)
0.481 mol Y = 0.481 mol H2O = 8.658 g
Total solution = 68.658 g.
Use density to convert this to volume.
Volume = mass/d = 66.879 mL.
M = mols/L = 1 mol/0.066879 = 14.95 M which probably should be rounded to 15 M since it appears we have only 3 significant figures.
Check my thinking. Check my work.

i understand everything you put there EXCEPT solving for x.

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1. i get how you get X = 2.077*Y, but i don't understand how if you plug in 0.481X in for Y to solve for x.

x = 2.077(0.481x)
x = 0.999x

then what???

i know this is probably a simple math question, but i need to do another problem like this and i'm stuck at this part...

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